Determinant of E = 1(1*0 - 4*6) - 2(0*0 - 4*5) + 3(0*6 - 1*5) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1.

Determinant of E = (1*4) - (2*2) = 4 - 4 = 0.

The discriminant must be non-negative: 2² - 4*1*k ≥ 0, which simplifies to k ≤ 1.

The discriminant must be less than zero: 6² - 4*1*k < 0, which simplifies to k > 9.

The discriminant must be non-negative: 6² - 4*1*k ≥ 0, thus k ≤ 9. The minimum value of k is -9.

The discriminant must be less than zero: k² - 4*1*9 < 0, thus k² < 36, so k > 6 or k < -6.

For both roots to be positive, k must be greater than 0 and k² > 4*9 = 36, thus k > 6.

For both roots to be negative, m must be greater than 0 and m² < 36. Thus, m must be in the range (-6, 0). The suitable value is -4.

The sum of the roots is 3 + 3 = 6, hence m = 6.

Using the sum of roots: 2 + (-2) = -p, hence p = 0.

For both roots to be negative, p must be greater than 0 and p² > 16. Thus, p < -4.

For both roots to be negative, p must be positive and p² > 4*9. Thus, p > 6, so p = -4 is valid.

The sum of the roots is 3 + 3 = 6, hence p = 6.

The first digit can be any of 10 digits, the second can be any of 9, the third can be any of 8, and the fourth can be any of 7. Total = 10 * 9 * 8 * 7 = 5040.

Each digit can be any of the 10 digits (0-9), so the total number of PINs is 10^4 = 10000.

The number of 4-digit PINs is P(10, 4) = 5040.

The number of ways to choose 2 from 5 is C(5, 2) = 10.

The number of ways to arrange 5 people is 5! = 120.

The number of ways to distribute 4 prizes among 10 students is 10P4 = 10! / (10-4)! = 5040.

The number of arrangements of 6 people is 6! = 720.

The number of arrangements of the letters in 'BOOK' is 4! / 2! = 12.

The number of arrangements of the letters in 'MATH' is 4! = 24.

The number of ways is 5C2 * 6C3 = 10 * 20 = 200.

The number of ways is 5C2 * 7C3 = 10 * 35 = 350.

The number of ways is 8C2 * 6C3 = 28 * 20 = 560.

The number of ways to select 2 boys from 5 is 5C2 and 3 girls from 7 is 7C3. Total = 5C2 * 7C3 = 10 * 35 = 350.

The number of ways to select 2 students from 5 is given by 5C2 = 10.

The number of ways to select 2 from 8 is C(8, 2) = 28.

The number of ways to select 3 books from 10 is given by 10C3 = 120.

The number of ways to arrange 5 people is 5! = 120.